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12 Nov 2024, 03:49
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x: 16
y: 3
A teacher distributes a batch of 56 cookies among \(x\) students such that each student receives exactly \(y\) cookies, where \(y > 0\), with some cookies left over. If these leftover cookies were redistributed, all except eight students would receive one additional cookie.
Select for x the possible value of \(x\), and select for y the possible value of \(y\) that would be jointly consistent with the given information. Make only two selections, one in each column.
Select for x the possible value of \(x\), and select for y the possible value of \(y\) that would be jointly consistent with the given information. Make only two selections, one in each column.
| x | y | |
| 2 | ||
| 3 | ||
| 4 | ||
| 15 | ||
| 16 | ||
| 32 |
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Official Answer
x: 16
y: 3
12 Nov 2024, 03:51
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Official Solution:
The number of cookies initially distributed was \(xy\).
Since redistributing the leftover cookies would give all except 8 students an additional cookie, the number of leftover cookies must have been \(x - 8\). This also indicates that there were more than 8 students in the group, so \(x > 8\).
Thus, we get \(xy + (x - 8) = 56\), which simplifies to \(x(y + 1) = 64\). This implies that \(x\) must be a factor of 64. The factors of 64 greater than 8 are 16, 32, and 64.
If \(x = 16\), then \(y = 3\).
If \(x = 32\), then \(y = 1\).
If \(x = 64\), then \(y = 0\), which is not possible since we are given that \(y > 0\).
Therefore, the possible pairs \((x, y)\) are \((16, 3)\) and \((32, 1)\). Among these pairs, only \((16, 3)\) is included in the options.
Correct answer:
x "16"
y "3"
Bunuel
The number of cookies initially distributed was \(xy\).
Since redistributing the leftover cookies would give all except 8 students an additional cookie, the number of leftover cookies must have been \(x - 8\). This also indicates that there were more than 8 students in the group, so \(x > 8\).
Thus, we get \(xy + (x - 8) = 56\), which simplifies to \(x(y + 1) = 64\). This implies that \(x\) must be a factor of 64. The factors of 64 greater than 8 are 16, 32, and 64.
If \(x = 16\), then \(y = 3\).
If \(x = 32\), then \(y = 1\).
If \(x = 64\), then \(y = 0\), which is not possible since we are given that \(y > 0\).
Therefore, the possible pairs \((x, y)\) are \((16, 3)\) and \((32, 1)\). Among these pairs, only \((16, 3)\) is included in the options.
Correct answer:
x "16"
y "3"
04 Jan 2025, 15:07
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I have revised the question, solution, and formatting by adding more details to enhance clarity.
